What is the units digit of the product of several numbers

Question

This problem is quite challenging to me. I know the answer should be an even number, but not know how to solve the problem. Thank you for help!

What is the units digit of the product $2^1·2^2· 2^3 · 2^4 · · · 2^{199}· 2^{200}$ where the numbers multiplied are the whole number powers from $2^1$ to $2^{200}$?

Answer 1

Hint: $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$$\prod_{i=1}^n2^i=2^{\sum_{i=1}^{200}i}$$

Use modulo arithmetic to determin the remainder after dividing with $10$ as this signifies the unit digit.

Answer 2

Actually, on second thought this is a perfect problem for one to tinker, think about, and come up with a reason on one's own and to try to figure out why. So I sort of regret giving the answer.

Before you read my answer try figuring things out on your own.

Look at $2^12^2...... 2^{200}$ and try

$2*4*8*16*32*64*128*.....*2^{200}$ and think about what you need (and more importantly what you don't need) to find the last digit.

Try to notice if there are any patterns in last digits. Try to figure out if you can see why the pattern happens.

Try smaller samples. What is the last digit of $2*4*8*16*32*64*128*256*512*1024$? If you only need the last digit do we really need to completely multiply out 256 or can we just use the part that will give us the last digit?

If there's a pattern see if you can lump this all into simpler things and why.

Then read my answer below.

======MY ANSWER: But muck around on your own before reading=======

$2^12^22^3....2^{200} = 2^{1+2+3+4...+ 200}$.

$1 + 2 + 3...... + 200 = 1/2[(1+2+3....+ 200) + (200 + 199 + .... 1)) = 1/2*200*201 = 20100$.

So what is the last digits of $2^{20100}$?

Notice that the last digit of $2^{n+5} = 32*2^n$ is the same last digit as $2*2^n = 2^{n+1}$. So the pattern of digits repeat every four interations.

Test:

$2^1 = 2 \implies 2$

$2^2 = 4 \implies 4$

$2^3 = 8 \implies 8$

$2^4 = 16 \implies 6$

$2^5 = 32 \implies 2$

$2^6 = 64 \implies 4$

.....

So $2^{20100}=2^{4*something}=2^{4*somethingminusone + 4}$ will have the same last digit as $2^4 = 16$ so the last digit is $6$.

==== or by mucking ====

$2^12^22^3.... 2^{200} = 2*4*8*16*32*64*128*.....$. Since we only want the last digit we can ignor everything but the last digits. e.g. $32*64 = (30+2)(60+4) = a Bunch Of Crap That Ends In Zero + 2*4$. So we can ignore the $3x$ and the $6x$ and only look at the $2$ and the $4$.

So $2*4*8*16*32*64*128*.... \approx 2*4*8*6*2*4*8....$

Hmm, that's interesting. We have a pattern 2,4,6,8 then 2,4,6,8 again. I wonder why.

Well once we get 32 that's equivalent to 2 so we just repeat.

So $2*4*8*6*2*4*8*6 ........ 2*4*8*6$.

Well $2*4*8*6 = 8*48 \approx 8*8 = 64 \approx 4$

So we have $4*4*4.....$ 50 times. $4*4 = 16 \implies 6$ so we havee $6*6*6$ 25 times. $6*6 = 36 = 6$ and $6*6*6*6 =6*6*6 = 6*6 = 6$ etc.

So $6*6*6... \implies 6$